Poisson and Binomial Questions

Date: 06/13/98 at 11:56:23
From: Kenneth W. Hodgkinson
Subject: Probability
1) Why does the waiting time for an event in a Poisson process exceed
t with the probability e^(-Lambda*t)?
2) Can a binomial variable can be thought of as a sum? Please explain
this, and use this idea to find the mean of a binomial variable.

Date: 06/13/98 at 17:04:04
From: Doctor Anthony
Subject: Re: Probability
Problem 1:
I will illustrate with an example. Suppose the number of accidents per
week on a certain stretch of road follows a Poisson distribution with
parameter 2.
If the average is 2 per week, then in time t weeks, the average would
be 2t.
The probability of no accident P(0) in time t is therefore e^(-2t)
The probability:
P(T>t) = probability that there are no events in time 0 to t
= e^(-2t)
P(T<t) = 1 - e^(-2t)
This is the cumulative distribution function (cdf) of the time to
first event.
To get the probability density function (pdf) we differentiate this
and get:
f(t) = 2e^(-2t)
This is pdf of time interval between events. It is called the
exponential distribution.
Problem 2:
Suppose we have probability p of success at each trial and probability
q of failure. The expectation in one trial is given by:
E(X) = p.1 + q.0 = p
Suppose now we have n trials, let X1, X2, X3, ......., Xn be the
results of these trials with any Xr taking values 0 or 1 with
probability q and p respectively.
If:
X = X1 + X2 + X3 + X4 + ...... + Xn
(= total successes in n trials)
then:
E(X) = E(X1) + E(X2) + E(X3) + .... + E(Xn)
= p + p + p + p + .........+ p
= np
So the mean of the binomial distribution with parametrs n and p is np.
-Doctor Anthony, The Math Forum
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